A) 2.14 V
B) 1.80 V
C) 1.07 V
D) 0.82 V
Correct Answer: C
Solution :
\[{{E}_{cell}}=E_{cell}^{o}-\frac{0.0591}{n}\log \,Q\] \[\underset{0.1\,M}{\mathop{C{{u}^{2+}}}}\,+Zn\xrightarrow{{}}\underset{1\,M}{\mathop{Z{{n}^{2+}}}}\,+Cu\] \[Q=\frac{[Z{{n}^{^{2+}}}]}{[C{{u}^{2+}}]}=\frac{1}{0.1}=10\] \[{{E}_{cell}}=1.10-\frac{0.0591}{2}\log 10\] \[=1.10-0.0295=1.0705\,V\]You need to login to perform this action.
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